What is the mass of water that results from combining 2.0 g of hydrogen with 16.0 g of oxygen?

Sodium metal (Na) reacts explosively with water to form sodium hydroxide (NaOH) and hydrogen gas. How many moles of sodium are n

Sati [7]

Answer:

308 moles of sodium

Explanation:

The balanced equation for the chemical reaction between sodium metal (Na) and water (H₂O) is the following:

2 Na(s) + 2 H₂O → 2 NaOH(aq) + H₂(g)

From the equation, we can see that 2 moles of Na react with 2 moles of H₂O to give 2 moles of NaOH and 1 mol of H₂ (hydrogen gas). So the stoichiometric mole ratio between Na and H₂ is: 2 mol Na/1 mol H₂. Thus, we multiply the mole ratio by the moles of H₂ to be produced to obtain the moles of Na required:

moles of Na required = 2 mol Na/1 mol H₂ x 154 moles H₂ = 308 moles Na

Therefore, 308 moles of sodium are needed to produce 154 moles of hydrogen gas.

7 0

2 years ago

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

3 years ago

Why is ozone an important form of oxygen?

Anton [14]

Ozone is made up of 3 atoms of Oxygen and present in stratosphere of our atmosphere. It prevents us from "Ultraviolet rays" of Sunlight so it is an important form of oxygen or us

Hope this helps!

4 0

2 years ago

The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20g/ml. the density of mercury is 13.6g/ml. part a what

valina [46]

As the atmospheric pressure is, P = dgh

Here d is the density of the mercury,

g is gravitation = 9.8 m/s²

h is height of the column, P = 751 torr = (751 torr × 1 atm / 760 torr) (101325 Pa) (1 N/m² / 1 Pa) = 100125 N/m²

Where, 1 N = 1 Kg / ms²

Thus, P = 100125 Kg / m³. s²

Therefore, height of the mercury column, when the atmospheric pressure is 751 torr,

h = P / gd

= (100125 kg / m³. s²) / (9.8 m/s²) (13.6 × 10³ kg / m³) = 0.751 m

As, d₁h₁ = d₂h₂

Here, d₁ is the density of the non-volatile liquid = 1.20 g/ml

d₂ is the density of the mercury = 13.6 g/ml

h₂ = 0.751 m

Thus, putting the values we get,

h₁ = d₂h₂ /d₁ = 13.6 g/ml × 0.751 m / 1.20 g/ml

= 8.5 m

3 0

2 years ago

Anybody can help? <br> - 20 points

strojnjashka [21]

Answer:

I think its carbon because of all the chemicals being used in that air.

7 0

2 years ago